A function in $W^{2,p}$ for $p>n/2$ is a.e. second differentiable
$begingroup$
Let $B$ be the unit ball in $mathbb{R}^{n}$ and $uin W^{2,p}(B)$, with $p>dfrac{n}{2}$. How can we see that $u$ is second differentiable almost everywhere in $B$?
This result is claimed in Page 25 but I cannot prove it. I can understand the proof of: $vin W^{1,p}(B)$, with $p>n$ then $v$ is differentiable almost everywhere in $B$ (in the same page of the link).
My attempt so far:
I try to follow the same method as in Page 25.
Assume that $0$ is a Lebesgue point of $D^2uin L^p$, i.e.
$$ |B_r|^{-1}int_{B_r}|D^2u(x)-D^2u(0)|^prightarrow 0, text{as }rrightarrow 0. quad (*)
$$
Here $B_r$ denotes the ball of radius $r$ centered at the origin. Now, our aim is to show that $u$ is classically second diffentiable at $0$. Set
$$
h(x)= u(x)-u(0)-Du(0)x-D^2u(0)dfrac{x^2}{2},
$$
and set $h_r(x):=h(rx)/r^{2}$. It is clear to see that it suffices to show
$$
||h_r||_{L^{infty}(B)}rightarrow 0, text{as }rrightarrow 0.quad textbf{(1)}
$$
Since $uin W^{2,p}(B)$, with $p>dfrac{n}{2}$, it is not hard to see that $uin W^{1,q}(B)$, for some $q>n$. For example, assume for simplicity that $n>p>frac{n}{2}$ then we can choose $q=np/(n-p)$.
By Morrey's inequality (since $q>n$),
$$
||h_r-h_r(0)||_{L^{infty}(B)}leq C ||Dh_r||_{L^q(B)}.
$$
Note that $h_r(0)=0$, and (*) is equivalent to $||D^2h_r||_{L^p(B)}rightarrow 0$. Therefore, in order to finish the proof (i.e. to show $textbf{(1)}$ is true), we really hope to have the following inequality
$$
||Dh_r||_{L^{q}(B)}leq C ||D^{2}h_{r}||_{L^{p}(B)}.
$$
However, this cannot be seen by the Sobolev embedding theorem since the Sobolev embedding on a bounded domain should be
$$
||Dh_r-|B|^{-1}int_{B}Dh_r||_{L^{q}(B)}leq C ||D^{2}h_{r}||_{L^{p}(B)}.
$$
How can I resolve this problem?
Thanks for any suggestion.
real-analysis sobolev-spaces regularity-theory-of-pdes
asked Dec 16 '18 at 16:39
HahnHahn
597
$endgroup$
add a comment |
$begingroup$
Let $B$ be the unit ball in $mathbb{R}^{n}$ and $uin W^{2,p}(B)$, with $p>dfrac{n}{2}$. How can we see that $u$ is second differentiable almost everywhere in $B$?
This result is claimed in Page 25 but I cannot prove it. I can understand the proof of: $vin W^{1,p}(B)$, with $p>n$ then $v$ is differentiable almost everywhere in $B$ (in the same page of the link).
My attempt so far:
I try to follow the same method as in Page 25.
Assume that $0$ is a Lebesgue point of $D^2uin L^p$, i.e.
$$ |B_r|^{-1}int_{B_r}|D^2u(x)-D^2u(0)|^prightarrow 0, text{as }rrightarrow 0. quad (*)
$$
Here $B_r$ denotes the ball of radius $r$ centered at the origin. Now, our aim is to show that $u$ is classically second diffentiable at $0$. Set
$$
h(x)= u(x)-u(0)-Du(0)x-D^2u(0)dfrac{x^2}{2},
$$
and set $h_r(x):=h(rx)/r^{2}$. It is clear to see that it suffices to show
$$
||h_r||_{L^{infty}(B)}rightarrow 0, text{as }rrightarrow 0.quad textbf{(1)}
$$
Since $uin W^{2,p}(B)$, with $p>dfrac{n}{2}$, it is not hard to see that $uin W^{1,q}(B)$, for some $q>n$. For example, assume for simplicity that $n>p>frac{n}{2}$ then we can choose $q=np/(n-p)$.
By Morrey's inequality (since $q>n$),
$$
||h_r-h_r(0)||_{L^{infty}(B)}leq C ||Dh_r||_{L^q(B)}.
$$
Note that $h_r(0)=0$, and (*) is equivalent to $||D^2h_r||_{L^p(B)}rightarrow 0$. Therefore, in order to finish the proof (i.e. to show $textbf{(1)}$ is true), we really hope to have the following inequality
$$
||Dh_r||_{L^{q}(B)}leq C ||D^{2}h_{r}||_{L^{p}(B)}.
$$
However, this cannot be seen by the Sobolev embedding theorem since the Sobolev embedding on a bounded domain should be
$$
||Dh_r-|B|^{-1}int_{B}Dh_r||_{L^{q}(B)}leq C ||D^{2}h_{r}||_{L^{p}(B)}.
$$
How can I resolve this problem?
Thanks for any suggestion.
real-analysis sobolev-spaces regularity-theory-of-pdes
asked Dec 16 '18 at 16:39
HahnHahn
597
$endgroup$
add a comment |
$begingroup$
Let $B$ be the unit ball in $mathbb{R}^{n}$ and $uin W^{2,p}(B)$, with $p>dfrac{n}{2}$. How can we see that $u$ is second differentiable almost everywhere in $B$?
This result is claimed in Page 25 but I cannot prove it. I can understand the proof of: $vin W^{1,p}(B)$, with $p>n$ then $v$ is differentiable almost everywhere in $B$ (in the same page of the link).
My attempt so far:
I try to follow the same method as in Page 25.
Assume that $0$ is a Lebesgue point of $D^2uin L^p$, i.e.
$$ |B_r|^{-1}int_{B_r}|D^2u(x)-D^2u(0)|^prightarrow 0, text{as }rrightarrow 0. quad (*)
$$
Here $B_r$ denotes the ball of radius $r$ centered at the origin. Now, our aim is to show that $u$ is classically second diffentiable at $0$. Set
$$
h(x)= u(x)-u(0)-Du(0)x-D^2u(0)dfrac{x^2}{2},
$$
and set $h_r(x):=h(rx)/r^{2}$. It is clear to see that it suffices to show
$$
||h_r||_{L^{infty}(B)}rightarrow 0, text{as }rrightarrow 0.quad textbf{(1)}
$$
Since $uin W^{2,p}(B)$, with $p>dfrac{n}{2}$, it is not hard to see that $uin W^{1,q}(B)$, for some $q>n$. For example, assume for simplicity that $n>p>frac{n}{2}$ then we can choose $q=np/(n-p)$.
By Morrey's inequality (since $q>n$),
$$
||h_r-h_r(0)||_{L^{infty}(B)}leq C ||Dh_r||_{L^q(B)}.
$$
Note that $h_r(0)=0$, and (*) is equivalent to $||D^2h_r||_{L^p(B)}rightarrow 0$. Therefore, in order to finish the proof (i.e. to show $textbf{(1)}$ is true), we really hope to have the following inequality
$$
||Dh_r||_{L^{q}(B)}leq C ||D^{2}h_{r}||_{L^{p}(B)}.
$$
However, this cannot be seen by the Sobolev embedding theorem since the Sobolev embedding on a bounded domain should be
$$
||Dh_r-|B|^{-1}int_{B}Dh_r||_{L^{q}(B)}leq C ||D^{2}h_{r}||_{L^{p}(B)}.
$$
How can I resolve this problem?
Thanks for any suggestion.
real-analysis sobolev-spaces regularity-theory-of-pdes
asked Dec 16 '18 at 16:39
HahnHahn
597
$endgroup$
Let $B$ be the unit ball in $mathbb{R}^{n}$ and $uin W^{2,p}(B)$, with $p>dfrac{n}{2}$. How can we see that $u$ is second differentiable almost everywhere in $B$?
This result is claimed in Page 25 but I cannot prove it. I can understand the proof of: $vin W^{1,p}(B)$, with $p>n$ then $v$ is differentiable almost everywhere in $B$ (in the same page of the link).
My attempt so far:
I try to follow the same method as in Page 25.
Assume that $0$ is a Lebesgue point of $D^2uin L^p$, i.e.
$$ |B_r|^{-1}int_{B_r}|D^2u(x)-D^2u(0)|^prightarrow 0, text{as }rrightarrow 0. quad (*)
$$
Here $B_r$ denotes the ball of radius $r$ centered at the origin. Now, our aim is to show that $u$ is classically second diffentiable at $0$. Set
$$
h(x)= u(x)-u(0)-Du(0)x-D^2u(0)dfrac{x^2}{2},
$$
and set $h_r(x):=h(rx)/r^{2}$. It is clear to see that it suffices to show
$$
||h_r||_{L^{infty}(B)}rightarrow 0, text{as }rrightarrow 0.quad textbf{(1)}
$$
Since $uin W^{2,p}(B)$, with $p>dfrac{n}{2}$, it is not hard to see that $uin W^{1,q}(B)$, for some $q>n$. For example, assume for simplicity that $n>p>frac{n}{2}$ then we can choose $q=np/(n-p)$.
By Morrey's inequality (since $q>n$),
$$
||h_r-h_r(0)||_{L^{infty}(B)}leq C ||Dh_r||_{L^q(B)}.
$$
Note that $h_r(0)=0$, and (*) is equivalent to $||D^2h_r||_{L^p(B)}rightarrow 0$. Therefore, in order to finish the proof (i.e. to show $textbf{(1)}$ is true), we really hope to have the following inequality
$$
||Dh_r||_{L^{q}(B)}leq C ||D^{2}h_{r}||_{L^{p}(B)}.
$$
However, this cannot be seen by the Sobolev embedding theorem since the Sobolev embedding on a bounded domain should be
$$
||Dh_r-|B|^{-1}int_{B}Dh_r||_{L^{q}(B)}leq C ||D^{2}h_{r}||_{L^{p}(B)}.
$$
How can I resolve this problem?
Thanks for any suggestion.
real-analysis sobolev-spaces regularity-theory-of-pdes
real-analysis sobolev-spaces regularity-theory-of-pdes
asked Dec 16 '18 at 16:39
HahnHahn
597
asked Dec 16 '18 at 16:39
HahnHahn
597
edited Dec 16 '18 at 17:30
Hahn
asked Dec 16 '18 at 16:39
HahnHahn
597
asked Dec 16 '18 at 16:39
HahnHahn
597
asked Dec 16 '18 at 16:39
HahnHahn
597
597
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: From
$$
left|Dh_r-|B|^{-1}int_{B}Dh_rright|_{L^{q}(B)}leq C |D^{2}h_{r}|_{L^{p}(B)}.
$$
and the triangle inequality, you have
$$
left|Dh_rright|_{L^{q}(B)}leq C |D^{2}h_{r}|_{L^{p}(B)} + hat C , left|int_{B}Dh_rright|.
$$
answered Dec 17 '18 at 7:17
gerwgerw
19.6k11334
$endgroup$
$begingroup$
Indeed, I still cannot see how $||Dh_r||_{L^{q}}rightarrow 0$. Are you talking about Holder's inequality to do more?
$endgroup$
– Hahn
Dec 17 '18 at 8:17
$begingroup$
With which term on the right-hand side do you have problems?
$endgroup$
– gerw
Dec 17 '18 at 8:32
$begingroup$
Since $||D^{2}h_{r}||_{L^{p}} rightarrow 0$, we need something like $ hat C , left|int_{B}Dh_rright|< (1-epsilon) ||Dh_{r}||_{L^{q}}$ to finish, right? But it's not clear. Note that, by applying Holder's inq, you would get $$ hat C , left|int_{B}Dh_rright| leq ||Dh_{r}||_{L^{q}}.$$
$endgroup$
– Hahn
Dec 17 '18 at 8:36
$begingroup$
If I am not mistaken, you have $int_B nabla h_r = r^{-n} , int_{B_r} nabla h = r^{-n} , int_{B_r} nabla u(x) - nabla u(0)$. If $0$ is a Lebesgue point of $nabla u$, this goes to zero.
$endgroup$
– gerw
Dec 17 '18 at 8:39
$begingroup$
No, it should be $int_B nabla h_r = r^{-n-1} , int_{B_r} nabla h $.
$endgroup$
– Hahn
Dec 17 '18 at 8:51
|
show 1 more comment
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$begingroup$
Hint: From
$$
left|Dh_r-|B|^{-1}int_{B}Dh_rright|_{L^{q}(B)}leq C |D^{2}h_{r}|_{L^{p}(B)}.
$$
and the triangle inequality, you have
$$
left|Dh_rright|_{L^{q}(B)}leq C |D^{2}h_{r}|_{L^{p}(B)} + hat C , left|int_{B}Dh_rright|.
$$
answered Dec 17 '18 at 7:17
gerwgerw
19.6k11334
$endgroup$
$begingroup$
Indeed, I still cannot see how $||Dh_r||_{L^{q}}rightarrow 0$. Are you talking about Holder's inequality to do more?
$endgroup$
– Hahn
Dec 17 '18 at 8:17
$begingroup$
With which term on the right-hand side do you have problems?
$endgroup$
– gerw
Dec 17 '18 at 8:32
$begingroup$
Since $||D^{2}h_{r}||_{L^{p}} rightarrow 0$, we need something like $ hat C , left|int_{B}Dh_rright|< (1-epsilon) ||Dh_{r}||_{L^{q}}$ to finish, right? But it's not clear. Note that, by applying Holder's inq, you would get $$ hat C , left|int_{B}Dh_rright| leq ||Dh_{r}||_{L^{q}}.$$
$endgroup$
– Hahn
Dec 17 '18 at 8:36
$begingroup$
If I am not mistaken, you have $int_B nabla h_r = r^{-n} , int_{B_r} nabla h = r^{-n} , int_{B_r} nabla u(x) - nabla u(0)$. If $0$ is a Lebesgue point of $nabla u$, this goes to zero.
$endgroup$
– gerw
Dec 17 '18 at 8:39
$begingroup$
No, it should be $int_B nabla h_r = r^{-n-1} , int_{B_r} nabla h $.
$endgroup$
– Hahn
Dec 17 '18 at 8:51
|
show 1 more comment
$begingroup$
Hint: From
$$
left|Dh_r-|B|^{-1}int_{B}Dh_rright|_{L^{q}(B)}leq C |D^{2}h_{r}|_{L^{p}(B)}.
$$
and the triangle inequality, you have
$$
left|Dh_rright|_{L^{q}(B)}leq C |D^{2}h_{r}|_{L^{p}(B)} + hat C , left|int_{B}Dh_rright|.
$$
answered Dec 17 '18 at 7:17
gerwgerw
19.6k11334
$endgroup$
$begingroup$
Indeed, I still cannot see how $||Dh_r||_{L^{q}}rightarrow 0$. Are you talking about Holder's inequality to do more?
$endgroup$
– Hahn
Dec 17 '18 at 8:17
$begingroup$
With which term on the right-hand side do you have problems?
$endgroup$
– gerw
Dec 17 '18 at 8:32
$begingroup$
Since $||D^{2}h_{r}||_{L^{p}} rightarrow 0$, we need something like $ hat C , left|int_{B}Dh_rright|< (1-epsilon) ||Dh_{r}||_{L^{q}}$ to finish, right? But it's not clear. Note that, by applying Holder's inq, you would get $$ hat C , left|int_{B}Dh_rright| leq ||Dh_{r}||_{L^{q}}.$$
$endgroup$
– Hahn
Dec 17 '18 at 8:36
$begingroup$
If I am not mistaken, you have $int_B nabla h_r = r^{-n} , int_{B_r} nabla h = r^{-n} , int_{B_r} nabla u(x) - nabla u(0)$. If $0$ is a Lebesgue point of $nabla u$, this goes to zero.
$endgroup$
– gerw
Dec 17 '18 at 8:39
$begingroup$
No, it should be $int_B nabla h_r = r^{-n-1} , int_{B_r} nabla h $.
$endgroup$
– Hahn
Dec 17 '18 at 8:51
|
show 1 more comment
$begingroup$
Hint: From
$$
left|Dh_r-|B|^{-1}int_{B}Dh_rright|_{L^{q}(B)}leq C |D^{2}h_{r}|_{L^{p}(B)}.
$$
and the triangle inequality, you have
$$
left|Dh_rright|_{L^{q}(B)}leq C |D^{2}h_{r}|_{L^{p}(B)} + hat C , left|int_{B}Dh_rright|.
$$
answered Dec 17 '18 at 7:17
gerwgerw
19.6k11334
$endgroup$
Hint: From
$$
left|Dh_r-|B|^{-1}int_{B}Dh_rright|_{L^{q}(B)}leq C |D^{2}h_{r}|_{L^{p}(B)}.
$$
and the triangle inequality, you have
$$
left|Dh_rright|_{L^{q}(B)}leq C |D^{2}h_{r}|_{L^{p}(B)} + hat C , left|int_{B}Dh_rright|.
$$
answered Dec 17 '18 at 7:17
gerwgerw
19.6k11334
answered Dec 17 '18 at 7:17
gerwgerw
19.6k11334
answered Dec 17 '18 at 7:17
gerwgerw
19.6k11334
answered Dec 17 '18 at 7:17
gerwgerw
19.6k11334
19.6k11334
$begingroup$
Indeed, I still cannot see how $||Dh_r||_{L^{q}}rightarrow 0$. Are you talking about Holder's inequality to do more?
$endgroup$
– Hahn
Dec 17 '18 at 8:17
$begingroup$
With which term on the right-hand side do you have problems?
$endgroup$
– gerw
Dec 17 '18 at 8:32
$begingroup$
Since $||D^{2}h_{r}||_{L^{p}} rightarrow 0$, we need something like $ hat C , left|int_{B}Dh_rright|< (1-epsilon) ||Dh_{r}||_{L^{q}}$ to finish, right? But it's not clear. Note that, by applying Holder's inq, you would get $$ hat C , left|int_{B}Dh_rright| leq ||Dh_{r}||_{L^{q}}.$$
$endgroup$
– Hahn
Dec 17 '18 at 8:36
$begingroup$
If I am not mistaken, you have $int_B nabla h_r = r^{-n} , int_{B_r} nabla h = r^{-n} , int_{B_r} nabla u(x) - nabla u(0)$. If $0$ is a Lebesgue point of $nabla u$, this goes to zero.
$endgroup$
– gerw
Dec 17 '18 at 8:39
$begingroup$
No, it should be $int_B nabla h_r = r^{-n-1} , int_{B_r} nabla h $.
$endgroup$
– Hahn
Dec 17 '18 at 8:51
|
show 1 more comment
$begingroup$
Indeed, I still cannot see how $||Dh_r||_{L^{q}}rightarrow 0$. Are you talking about Holder's inequality to do more?
$endgroup$
– Hahn
Dec 17 '18 at 8:17
$begingroup$
With which term on the right-hand side do you have problems?
$endgroup$
– gerw
Dec 17 '18 at 8:32
$begingroup$
Since $||D^{2}h_{r}||_{L^{p}} rightarrow 0$, we need something like $ hat C , left|int_{B}Dh_rright|< (1-epsilon) ||Dh_{r}||_{L^{q}}$ to finish, right? But it's not clear. Note that, by applying Holder's inq, you would get $$ hat C , left|int_{B}Dh_rright| leq ||Dh_{r}||_{L^{q}}.$$
$endgroup$
– Hahn
Dec 17 '18 at 8:36
$begingroup$
If I am not mistaken, you have $int_B nabla h_r = r^{-n} , int_{B_r} nabla h = r^{-n} , int_{B_r} nabla u(x) - nabla u(0)$. If $0$ is a Lebesgue point of $nabla u$, this goes to zero.
$endgroup$
– gerw
Dec 17 '18 at 8:39
$begingroup$
No, it should be $int_B nabla h_r = r^{-n-1} , int_{B_r} nabla h $.
$endgroup$
– Hahn
Dec 17 '18 at 8:51
$begingroup$
Indeed, I still cannot see how $||Dh_r||_{L^{q}}rightarrow 0$. Are you talking about Holder's inequality to do more?
$endgroup$
– Hahn
Dec 17 '18 at 8:17
$begingroup$
Indeed, I still cannot see how $||Dh_r||_{L^{q}}rightarrow 0$. Are you talking about Holder's inequality to do more?
$endgroup$
– Hahn
Dec 17 '18 at 8:17
$begingroup$
With which term on the right-hand side do you have problems?
$endgroup$
– gerw
Dec 17 '18 at 8:32
$begingroup$
With which term on the right-hand side do you have problems?
$endgroup$
– gerw
Dec 17 '18 at 8:32
$begingroup$
Since $||D^{2}h_{r}||_{L^{p}} rightarrow 0$, we need something like $ hat C , left|int_{B}Dh_rright|< (1-epsilon) ||Dh_{r}||_{L^{q}}$ to finish, right? But it's not clear. Note that, by applying Holder's inq, you would get $$ hat C , left|int_{B}Dh_rright| leq ||Dh_{r}||_{L^{q}}.$$
$endgroup$
– Hahn
Dec 17 '18 at 8:36
$begingroup$
Since $||D^{2}h_{r}||_{L^{p}} rightarrow 0$, we need something like $ hat C , left|int_{B}Dh_rright|< (1-epsilon) ||Dh_{r}||_{L^{q}}$ to finish, right? But it's not clear. Note that, by applying Holder's inq, you would get $$ hat C , left|int_{B}Dh_rright| leq ||Dh_{r}||_{L^{q}}.$$
$endgroup$
– Hahn
Dec 17 '18 at 8:36
$begingroup$
If I am not mistaken, you have $int_B nabla h_r = r^{-n} , int_{B_r} nabla h = r^{-n} , int_{B_r} nabla u(x) - nabla u(0)$. If $0$ is a Lebesgue point of $nabla u$, this goes to zero.
$endgroup$
– gerw
Dec 17 '18 at 8:39
$begingroup$
If I am not mistaken, you have $int_B nabla h_r = r^{-n} , int_{B_r} nabla h = r^{-n} , int_{B_r} nabla u(x) - nabla u(0)$. If $0$ is a Lebesgue point of $nabla u$, this goes to zero.
$endgroup$
– gerw
Dec 17 '18 at 8:39
$begingroup$
No, it should be $int_B nabla h_r = r^{-n-1} , int_{B_r} nabla h $.
$endgroup$
– Hahn
Dec 17 '18 at 8:51
$begingroup$
No, it should be $int_B nabla h_r = r^{-n-1} , int_{B_r} nabla h $.
$endgroup$
– Hahn
Dec 17 '18 at 8:51
|
show 1 more comment
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
